Problem 31
Question
Solve the rational inequality. $$\frac{x(x+1)}{1+x^{2}} \geq 0$$
Step-by-Step Solution
Verified Answer
The solution to the inequality is \([-1,0] ∪ (-∞,-1) ∪ (0,∞)\].
1Step 1: Identify Critical Points
To identify critical points, set both the numerator and denominator equal to zero and solve. For the numerator \(x(x+1) = 0\), the solutions are \(x = 0\) and \(x = -1\). For the denominator \(1+x^{2} = 0\), there is no real solution hence will not affect the sign of the function. Discontinuities are also where the denominator equals 0 but in this case, there are none.
2Step 2: Determine Sign Changes
The important intervals are \((-∞,-1)\), \((-1,0)\), and \((0,∞)\). Next, pick a test point in each interval and determine whether it makes the function positive or negative. For \((-∞,-1)\), \(-2\) might be a good test point, for \((-1,0)\), \(-0.5\) might be a good test point, and for \((0,∞)\), \(1\) is a good test point.
3Step 3: Evaluate Test Points
Plug the test points into the function: \((-2)[-2+1]/[1+(-2)^{2}] = (2)/[5] > 0\), \((-0.5)[-0.5+1]/[1+(-0.5)^{2}] = (0.5)/[1.25] > 0\), \(1[1+1]/[1+(1)^{2}] = (2)/[2] = 1 > 0\). In each interval, the function is positive.
4Step 4: Write The Final Answer
Since the inequality includes 'equals', we need to include the points \(x = -1\) and \(x = 0\) in the solution. Therefore, the solution to the inequality is \([-1,0] ∪ (-∞,-1) ∪ (0,∞)\]. This means \(x\) can be any real number.
Key Concepts
Critical PointsSign AnalysisNumerator and DenominatorSolutions of Inequalities
Critical Points
When solving rational inequalities, identifying critical points is an essential first step. Critical points occur where the numerator or denominator of the rational expression are equal to zero. These points are significant because they can indicate where the function changes its sign—shifting between positive and negative values—or where it might be undefined.
In this inequality, we first explore the numerator, which is a key component because it directly affects when the function equals zero or changes sign. The numerator, represented by the expression \(x(x+1)\), is zero at \(x = 0\) and \(x = -1\). These values point us to potential critical points. However, while determining the denominator \(1+ x^2\), we notice there are no real solutions, since \(x^2\) is always positive, and thus it never contributes to zeroing the expression.
Understanding the critical points allows us to segment the number line into intervals where we can test the behavior of the function.
In this inequality, we first explore the numerator, which is a key component because it directly affects when the function equals zero or changes sign. The numerator, represented by the expression \(x(x+1)\), is zero at \(x = 0\) and \(x = -1\). These values point us to potential critical points. However, while determining the denominator \(1+ x^2\), we notice there are no real solutions, since \(x^2\) is always positive, and thus it never contributes to zeroing the expression.
Understanding the critical points allows us to segment the number line into intervals where we can test the behavior of the function.
Sign Analysis
After identifying critical points, the next step is conducting a sign analysis over the intervals formed around these points. Sign analysis helps us understand where the expression is positive or negative. This is crucial in determining the solution set of the inequality.
For the inequality \(\frac{x(x+1)}{1+x^2} \geq 0\), the number line breaks into three intervals based on the critical points: \((-∞,-1)\), \((-1,0)\), and \((0,∞)\). Select a point within each interval and substitute back into the function. This helps ascertain the nature of the function over that interval:
For the inequality \(\frac{x(x+1)}{1+x^2} \geq 0\), the number line breaks into three intervals based on the critical points: \((-∞,-1)\), \((-1,0)\), and \((0,∞)\). Select a point within each interval and substitute back into the function. This helps ascertain the nature of the function over that interval:
- For \((-∞,-1)\), using \(x = -2\), the expression is positive.
- For \((-1,0)\), using \(x = -0.5\), the expression too is positive.
- For \((0,∞)\), using \(x = 1\), the expression remains positive.
Numerator and Denominator
In rational expressions and inequalities, the numerator and the denominator both play key roles. Understanding how each one influences the overall expression helps us solve inequalities efficiently.
The numerator in our inequality, \(x(x+1)\), is a polynomial that has zeros at \(x = 0\) and \(x = -1\). These zeros are also known as roots and are integral to finding critical points. In itself, the numerator being zero is valuable since it signifies where the rational expression could potentially equal zero, a scenario allowed by our non-strict inequality.
The denominator, \(1+x^2\), does not become zero because \(x^2\) is always non-negative, and adding 1 ensures a positive value. Thus, no real \(x\) can lead to a zero denominator, avoiding any undefined conditions. This steady insight from the denominator means our function remains continuous without interruptions, simplifying our sign analysis and solution.
The numerator in our inequality, \(x(x+1)\), is a polynomial that has zeros at \(x = 0\) and \(x = -1\). These zeros are also known as roots and are integral to finding critical points. In itself, the numerator being zero is valuable since it signifies where the rational expression could potentially equal zero, a scenario allowed by our non-strict inequality.
The denominator, \(1+x^2\), does not become zero because \(x^2\) is always non-negative, and adding 1 ensures a positive value. Thus, no real \(x\) can lead to a zero denominator, avoiding any undefined conditions. This steady insight from the denominator means our function remains continuous without interruptions, simplifying our sign analysis and solution.
Solutions of Inequalities
Finding the solutions to rational inequalities involves combining all the analysis performed on critical points, sign changes, and the behavior of the numerator and denominator. The solution set is framed by areas of the number line where the inequality holds true.
In the example \(\frac{x(x+1)}{1+x^2} \geq 0\), our analysis has shown the function is positive for three distinct intervals: \((-∞,-1)\), \((-1,0)\), and \((0,∞)\). Since our inequality is non-strict (includes the equal sign), we also include the points where the function equals zero (\(x = -1\) and \(x = 0\)). The union of these intervals covers all real numbers.
In the example \(\frac{x(x+1)}{1+x^2} \geq 0\), our analysis has shown the function is positive for three distinct intervals: \((-∞,-1)\), \((-1,0)\), and \((0,∞)\). Since our inequality is non-strict (includes the equal sign), we also include the points where the function equals zero (\(x = -1\) and \(x = 0\)). The union of these intervals covers all real numbers.
- The solution is expressed as \([-1,0] \cup (-∞,-1) \cup (0,∞)\).
- This implies any real number \(x\) satisfies the inequality, indicating a broad solution set capturing all real numbers.
Other exercises in this chapter
Problem 30
Determine the end behavior of the function. $$f(x)=3 x^{3}-4 x^{2}+5$$
View solution Problem 30
For each polynomial function, find (a) the end behavior; (b) the \(y\) -intercept; (c) the \(x\) -intercept(s) of the graph of the function and the multipliciti
View solution Problem 31
Find all the real zeros of the polynomial. $$P(x)=7 x^{3}+2 x^{2}-28 x-8$$
View solution Problem 31
Sketch a graph of the rational function. Indicate any vertical and horizontal asymptote(s) and all intercepts. $$f(x)=\frac{1}{x-2}$$
View solution